Why does a longer hydrocarbon usually have a more negative enthalpy of combustion?

In IB Chemistry, longer-chain hydrocarbons contain more carbon and hydrogen atoms, which means more bonds that can be oxidized. When the hydrocarbon combusts completely, it forms more moles of CO2 and H2O. Forming the C=O and O-H bonds releases a lot of energy, so the total energy released per mole of fuel tends to increase with chain length. That shows up as a more negative ΔHc° value. However, exams may also ask you to compare energy per gram, not per mole, so you should be ready to convert using molar mass. In short: more product bonds formed usually means more energy released, but check what “per” unit the question is using.

Why must combustion be complete for ΔHc° values?

The definition of standard enthalpy of combustion in IB Chemistry requires the fuel to be completely burned in oxygen. If combustion is incomplete, you can form CO(g) or even solid carbon (soot), which means the products are not the standard CO2(g) and H2O(l) set. Incomplete combustion releases less energy because the oxidation is not taken all the way to the most stable oxidation products. That is why experimental calorimetry often gives values that are less negative than the data booklet. In exams, “complete combustion” is also a hint: your equation must end in CO2 and H2O with correct balancing. If you miss that, your whole calculation chain can be consistent but still marked wrong.

Why is water written as liquid in enthalpy of combustion questions?

Under standard conditions used in IB Chemistry (298 K and 100 kPa), the standard state of water is liquid, H2O(l). That means ΔHc° values are defined with liquid water as a product, even if a real flame produces steam near the reaction zone. This detail matters because changing state involves enthalpy, and H2O(g) vs H2O(l) would change the energy value. Examiners like this because it tests whether you understand what “standard state” means, not just balancing equations. When you use the data booklet combustion table, it is aligned to H2O(l) for the same reason. So, unless the question explicitly says otherwise, assume H2O(l) and move on confidently.